-0.000 282 006 002 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 006 002(10) to 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

What are the steps to convert decimal number
-0.000 282 006 002(10) to 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. Start with the positive version of the number:

|-0.000 282 006 002| = 0.000 282 006 002


2. First, convert to binary (in base 2) the integer part: 0.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 0 ÷ 2 = 0 + 0;

3. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.000 282 006 002.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.000 282 006 002 × 2 = 0 + 0.000 564 012 004;
  • 2) 0.000 564 012 004 × 2 = 0 + 0.001 128 024 008;
  • 3) 0.001 128 024 008 × 2 = 0 + 0.002 256 048 016;
  • 4) 0.002 256 048 016 × 2 = 0 + 0.004 512 096 032;
  • 5) 0.004 512 096 032 × 2 = 0 + 0.009 024 192 064;
  • 6) 0.009 024 192 064 × 2 = 0 + 0.018 048 384 128;
  • 7) 0.018 048 384 128 × 2 = 0 + 0.036 096 768 256;
  • 8) 0.036 096 768 256 × 2 = 0 + 0.072 193 536 512;
  • 9) 0.072 193 536 512 × 2 = 0 + 0.144 387 073 024;
  • 10) 0.144 387 073 024 × 2 = 0 + 0.288 774 146 048;
  • 11) 0.288 774 146 048 × 2 = 0 + 0.577 548 292 096;
  • 12) 0.577 548 292 096 × 2 = 1 + 0.155 096 584 192;
  • 13) 0.155 096 584 192 × 2 = 0 + 0.310 193 168 384;
  • 14) 0.310 193 168 384 × 2 = 0 + 0.620 386 336 768;
  • 15) 0.620 386 336 768 × 2 = 1 + 0.240 772 673 536;
  • 16) 0.240 772 673 536 × 2 = 0 + 0.481 545 347 072;
  • 17) 0.481 545 347 072 × 2 = 0 + 0.963 090 694 144;
  • 18) 0.963 090 694 144 × 2 = 1 + 0.926 181 388 288;
  • 19) 0.926 181 388 288 × 2 = 1 + 0.852 362 776 576;
  • 20) 0.852 362 776 576 × 2 = 1 + 0.704 725 553 152;
  • 21) 0.704 725 553 152 × 2 = 1 + 0.409 451 106 304;
  • 22) 0.409 451 106 304 × 2 = 0 + 0.818 902 212 608;
  • 23) 0.818 902 212 608 × 2 = 1 + 0.637 804 425 216;
  • 24) 0.637 804 425 216 × 2 = 1 + 0.275 608 850 432;
  • 25) 0.275 608 850 432 × 2 = 0 + 0.551 217 700 864;
  • 26) 0.551 217 700 864 × 2 = 1 + 0.102 435 401 728;
  • 27) 0.102 435 401 728 × 2 = 0 + 0.204 870 803 456;
  • 28) 0.204 870 803 456 × 2 = 0 + 0.409 741 606 912;
  • 29) 0.409 741 606 912 × 2 = 0 + 0.819 483 213 824;
  • 30) 0.819 483 213 824 × 2 = 1 + 0.638 966 427 648;
  • 31) 0.638 966 427 648 × 2 = 1 + 0.277 932 855 296;
  • 32) 0.277 932 855 296 × 2 = 0 + 0.555 865 710 592;
  • 33) 0.555 865 710 592 × 2 = 1 + 0.111 731 421 184;
  • 34) 0.111 731 421 184 × 2 = 0 + 0.223 462 842 368;
  • 35) 0.223 462 842 368 × 2 = 0 + 0.446 925 684 736;
  • 36) 0.446 925 684 736 × 2 = 0 + 0.893 851 369 472;
  • 37) 0.893 851 369 472 × 2 = 1 + 0.787 702 738 944;
  • 38) 0.787 702 738 944 × 2 = 1 + 0.575 405 477 888;
  • 39) 0.575 405 477 888 × 2 = 1 + 0.150 810 955 776;
  • 40) 0.150 810 955 776 × 2 = 0 + 0.301 621 911 552;
  • 41) 0.301 621 911 552 × 2 = 0 + 0.603 243 823 104;
  • 42) 0.603 243 823 104 × 2 = 1 + 0.206 487 646 208;
  • 43) 0.206 487 646 208 × 2 = 0 + 0.412 975 292 416;
  • 44) 0.412 975 292 416 × 2 = 0 + 0.825 950 584 832;
  • 45) 0.825 950 584 832 × 2 = 1 + 0.651 901 169 664;
  • 46) 0.651 901 169 664 × 2 = 1 + 0.303 802 339 328;
  • 47) 0.303 802 339 328 × 2 = 0 + 0.607 604 678 656;
  • 48) 0.607 604 678 656 × 2 = 1 + 0.215 209 357 312;
  • 49) 0.215 209 357 312 × 2 = 0 + 0.430 418 714 624;
  • 50) 0.430 418 714 624 × 2 = 0 + 0.860 837 429 248;
  • 51) 0.860 837 429 248 × 2 = 1 + 0.721 674 858 496;
  • 52) 0.721 674 858 496 × 2 = 1 + 0.443 349 716 992;
  • 53) 0.443 349 716 992 × 2 = 0 + 0.886 699 433 984;
  • 54) 0.886 699 433 984 × 2 = 1 + 0.773 398 867 968;
  • 55) 0.773 398 867 968 × 2 = 1 + 0.546 797 735 936;
  • 56) 0.546 797 735 936 × 2 = 1 + 0.093 595 471 872;
  • 57) 0.093 595 471 872 × 2 = 0 + 0.187 190 943 744;
  • 58) 0.187 190 943 744 × 2 = 0 + 0.374 381 887 488;
  • 59) 0.374 381 887 488 × 2 = 0 + 0.748 763 774 976;
  • 60) 0.748 763 774 976 × 2 = 1 + 0.497 527 549 952;
  • 61) 0.497 527 549 952 × 2 = 0 + 0.995 055 099 904;
  • 62) 0.995 055 099 904 × 2 = 1 + 0.990 110 199 808;
  • 63) 0.990 110 199 808 × 2 = 1 + 0.980 220 399 616;
  • 64) 0.980 220 399 616 × 2 = 1 + 0.960 440 799 232;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (Losing precision - the converted number we get in the end will be just a very good approximation of the initial one).


5. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.000 282 006 002(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1000 1110 0100 1101 0011 0111 0001 0111(2)

6. Positive number before normalization:

0.000 282 006 002(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1000 1110 0100 1101 0011 0111 0001 0111(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 12 positions to the right, so that only one non zero digit remains to the left of it:


0.000 282 006 002(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1000 1110 0100 1101 0011 0111 0001 0111(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1000 1110 0100 1101 0011 0111 0001 0111(2) × 20 =

1.0010 0111 1011 0100 0110 1000 1110 0100 1101 0011 0111 0001 0111(2) × 2-12


8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

Sign 1 (a negative number)

Exponent (unadjusted): -12

Mantissa (not normalized):
1.0010 0111 1011 0100 0110 1000 1110 0100 1101 0011 0111 0001 0111


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-12 + 2(11-1) - 1 =

(-12 + 1 023)(10) =

1 011(10)


10. Convert the adjusted exponent from the decimal (base 10) to 11 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 1 011 ÷ 2 = 505 + 1;
  • 505 ÷ 2 = 252 + 1;
  • 252 ÷ 2 = 126 + 0;
  • 126 ÷ 2 = 63 + 0;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1011(10) =

011 1111 0011(2)


12. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 52 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 0010 0111 1011 0100 0110 1000 1110 0100 1101 0011 0111 0001 0111 =

0010 0111 1011 0100 0110 1000 1110 0100 1101 0011 0111 0001 0111


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (11 bits) =
011 1111 0011

Mantissa (52 bits) =
0010 0111 1011 0100 0110 1000 1110 0100 1101 0011 0111 0001 0111


Decimal number -0.000 282 006 002 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 1000 1110 0100 1101 0011 0111 0001 0111

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How to convert numbers from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 64 bit double precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders from the previous operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the multiplying operations, starting from the top of the list constructed above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, shifting the decimal mark (the decimal point) "n" positions either to the left, or to the right, so that only one non zero digit remains to the left of the decimal mark.
  • 7. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal mark, if the case) and adjust its length to 52 bits, either by removing the excess bits from the right (losing precision...) or by adding extra bits set on '0' to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -31.640 215 from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-31.640 215| = 31.640 215

  • 2. First convert the integer part, 31. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 31 ÷ 2 = 15 + 1;
    • 15 ÷ 2 = 7 + 1;
    • 7 ÷ 2 = 3 + 1;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    31(10) = 1 1111(2)

  • 4. Then, convert the fractional part, 0.640 215. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.640 215 × 2 = 1 + 0.280 43;
    • 2) 0.280 43 × 2 = 0 + 0.560 86;
    • 3) 0.560 86 × 2 = 1 + 0.121 72;
    • 4) 0.121 72 × 2 = 0 + 0.243 44;
    • 5) 0.243 44 × 2 = 0 + 0.486 88;
    • 6) 0.486 88 × 2 = 0 + 0.973 76;
    • 7) 0.973 76 × 2 = 1 + 0.947 52;
    • 8) 0.947 52 × 2 = 1 + 0.895 04;
    • 9) 0.895 04 × 2 = 1 + 0.790 08;
    • 10) 0.790 08 × 2 = 1 + 0.580 16;
    • 11) 0.580 16 × 2 = 1 + 0.160 32;
    • 12) 0.160 32 × 2 = 0 + 0.320 64;
    • 13) 0.320 64 × 2 = 0 + 0.641 28;
    • 14) 0.641 28 × 2 = 1 + 0.282 56;
    • 15) 0.282 56 × 2 = 0 + 0.565 12;
    • 16) 0.565 12 × 2 = 1 + 0.130 24;
    • 17) 0.130 24 × 2 = 0 + 0.260 48;
    • 18) 0.260 48 × 2 = 0 + 0.520 96;
    • 19) 0.520 96 × 2 = 1 + 0.041 92;
    • 20) 0.041 92 × 2 = 0 + 0.083 84;
    • 21) 0.083 84 × 2 = 0 + 0.167 68;
    • 22) 0.167 68 × 2 = 0 + 0.335 36;
    • 23) 0.335 36 × 2 = 0 + 0.670 72;
    • 24) 0.670 72 × 2 = 1 + 0.341 44;
    • 25) 0.341 44 × 2 = 0 + 0.682 88;
    • 26) 0.682 88 × 2 = 1 + 0.365 76;
    • 27) 0.365 76 × 2 = 0 + 0.731 52;
    • 28) 0.731 52 × 2 = 1 + 0.463 04;
    • 29) 0.463 04 × 2 = 0 + 0.926 08;
    • 30) 0.926 08 × 2 = 1 + 0.852 16;
    • 31) 0.852 16 × 2 = 1 + 0.704 32;
    • 32) 0.704 32 × 2 = 1 + 0.408 64;
    • 33) 0.408 64 × 2 = 0 + 0.817 28;
    • 34) 0.817 28 × 2 = 1 + 0.634 56;
    • 35) 0.634 56 × 2 = 1 + 0.269 12;
    • 36) 0.269 12 × 2 = 0 + 0.538 24;
    • 37) 0.538 24 × 2 = 1 + 0.076 48;
    • 38) 0.076 48 × 2 = 0 + 0.152 96;
    • 39) 0.152 96 × 2 = 0 + 0.305 92;
    • 40) 0.305 92 × 2 = 0 + 0.611 84;
    • 41) 0.611 84 × 2 = 1 + 0.223 68;
    • 42) 0.223 68 × 2 = 0 + 0.447 36;
    • 43) 0.447 36 × 2 = 0 + 0.894 72;
    • 44) 0.894 72 × 2 = 1 + 0.789 44;
    • 45) 0.789 44 × 2 = 1 + 0.578 88;
    • 46) 0.578 88 × 2 = 1 + 0.157 76;
    • 47) 0.157 76 × 2 = 0 + 0.315 52;
    • 48) 0.315 52 × 2 = 0 + 0.631 04;
    • 49) 0.631 04 × 2 = 1 + 0.262 08;
    • 50) 0.262 08 × 2 = 0 + 0.524 16;
    • 51) 0.524 16 × 2 = 1 + 0.048 32;
    • 52) 0.048 32 × 2 = 0 + 0.096 64;
    • 53) 0.096 64 × 2 = 0 + 0.193 28;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 52) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.640 215(10) = 0.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 6. Summarizing - the positive number before normalization:

    31.640 215(10) = 1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 7. Normalize the binary representation of the number, shifting the decimal mark 4 positions to the left so that only one non-zero digit stays to the left of the decimal mark:

    31.640 215(10) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 20 =
    1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

  • 9. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as shown above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1 = (4 + 1023)(10) = 1027(10) =
    100 0000 0011(2)

  • 10. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign) and adjust its length to 52 bits, by removing the excess bits, from the right (losing precision...):

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

    Mantissa (normalized): 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 100 0000 0011

    Mantissa (52 bits) = 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Number -31.640 215, converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point =
    1 - 100 0000 0011 - 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100